Book chapter ; Quantum technology, quantum computing, quantum sensing, quantum cryptography

1. 10. The electromagnetic field
10.1 Classical theory of the e.m. field
10.2 Quantization of the e.m. field
10.2.1 Zero-Point Energy and the Casimir Force
10.3 Quantization of the e.m. field in the Coulomb gauge
10.4 States of the e.m. field
10.4.1 Photon number eigenstates
10.4.2 Coherent states
10.4.3 Measurement Statistics
10.5 Atomic interactions with the quantized field
We will now provide a quanto-mechanical description of the electro-magnetic field. Our main interest will be in
analyzing phenomena linked to atomic physics and quantum optics, in which atoms interacts with radiation. Some
processes can be analyzed with a classical description: for example we studied the precessing and the manipulation
of a spin by classical static and rf magnetic fields. Absorption and emission of light by an atom can also be described
as the interaction with a classical field. Some other phenomena, such as spontaneous emission, can only arise from a
QM description of both the atom and the field. There are various examples in which the importance of a quantum
treatment of electromagnetism becomes evident:
– Casimir force
– Spontaneous emission, Lamb Shift
– Laser linewidth, photon statistics
– Squeezed photon states, states with subpoissonian distribution,
– Quantum beats, two photon interference, etc.
10.1 Classical theory of the e.m. field
Before introducing the quantization of the field, we want to review some basic (and relevant) concepts about e.m.
fields.
Maxwell equations for the electric and magnetic fields, E and B, are:
Gauss’s law ∇ · E = ρ
ε0
Gauss’s law for magnetism ∇ · B = 0
Maxwell-Faraday equation (Faraday’s law of induction) ∇ × E = −1 ∂B
c ∂t
Ampere’s circuital law (with Maxwell’s correction) ∇ × B = µ0J + µ0ε ∂E
0 ∂t
We will be interested to their solution in empty space (and setting c = 1/
√
µ0ε0):
Gauss’s law E = 0
Gauss’s law for magnetism
∇ ·
∇ · B = 0
Maxwell-Faraday equation ∇ × E = −1 ∂B
c ∂t
Ampere’s circuital law ∇ × B = 1 ∂E
c ∂t
93

2. Combining Maxwell equation in vacuum, we find the wave equations:
2 1 ∂2
E
∇ E − = 0 B
c2 t2
∇2 1 ∂2
B
∂
− = 0
c2 ∂t2
? Question: Show how this equation is derived
We need to take the curl of Maxwell-Faraday equation and the time derivative of Ampere’s law and use the vector identity
∇ × (∇ × ~
v) = ∇(∇ · ~
v) − ∇2
~
v and Gauss law.
~
A general solution for these equations can be written simply as E = E(ωt k ~
x). By fixing the boundary conditions,
we can find a solution in terms of an expansion in normal modes, where th
−
e tim
·
e dependence and spatial dependence
are separated. The solution of the wave equation can thus be facilitated by representing the electric field as a sum of
normal mode functions:
E(~
x, t) = fm(t)~
um(~
x).
m
The normal modes u are the equivalent of eigenfunc
X
m tions for the wave equation, so they do not evolve in time
(i.e. they are function of position only). The um are orthonormal functions, called normal modes. The boundary
conditions define the normal modes um for the field, satisfying:
∇2
um = −k2
mum, ∇ · um = 0, ~
n × um = 0
(where n is a unit vector normal to a surface). This last condition is imposed because the tangential component of
the electric field E must vanish on a conducting surface. We can also choose the modes to satisfy the orthonomality
condition (hence normal modes): Z
~
um(x)~
un(x)d3
x = δn,m
Substituting the expression for the electric field in the wave equation, we find an equation for the coefficient fm(t):
X d2
fm
+ c2
k2
mfm(t) = 0.
dt2
m
Since the mode functions are linearly independent, the coefficients of each mode must separately add up to zero in
order to satisfy the wave equation, and we find :
d2
fm
+ c2
k2
dt2 mfm(t) = 0.
As it can be seen from this equation, the dynamics of the normal modes, as described by their time-dependent
coefficients, is the same as that of the h.o. with frequency ωm = ckm. Hence the electric field is equivalent to an
infinite number of (independent) harmonic oscillators. In order to find a quantum-mechanical description of the e.m.
field we will need to turn this h.o. into quantum harmonic oscillators.
We want to express the magnetic field in terms of the same normal modes ~
um which are our basis. We assume for B
the expansion:
B(x, t) = hn(t) [
n
∇ × un(x)] ,
From Maxwell-Faraday law:
X
∇ × E =
X 1
fn(t)∇ × un = ∂
c
n
− tB
we see that we need to impose hn such that d hn
= −cfn so that we obtain
d t
X 1 d h
n
−
n 1
c d t
∇ × un = − ∂tB
c
94

3. which indeed corresponds to the desired expression for the magnetic field. We now want to find as well an equation
for the coefficient hn alone. From the expressions of the E and B-field in terms of normal modes, using Ampere’s
law,
1 ∂E 1 d f
∇ × B = →
X n
hn(t)∇ × (∇ × un) =
X
un
c ∂t c d t
n n
→ − hn∇2 1 d fn
un = un
c d t
n n
(where we used the fact that
X X
∇ · u = 0) we find
dfn(t)
= ck2
nhn(t).
dt
since we have ∇2
un = −k2
nun. Finally we have:
d2
hn(t) + c2
k2
hn(t) = 0
dt2 n
The Hamiltonian of the system represent the total energy33 2
P
: H = 1 1
(E + B2
)d3
x.
2 4π
We can show that H = 1 2 2 2
n (fn + k
8 nh
π n):
R
1
H =
Z
(E2
+ B2
)d3 1
x =
X
fnfm
Z
u x )d3
x + 3
n( )um(x hnhm
π 8π n,m
Z
(∇ × un) · (∇ × um)d x
8
=
X 1
(f2
+
8π n k2
nh2
n)
n
where we used
R
(∇ × f
un) ( m)d3
x = k2 d (t)
u nδn,m. We can then use the equation n
= ck2
nhn(t) to eliminate
dt
hn.Then fn can be associat
·
ed
∇
w
×
ith an equivalent position operator and hn (being a derivative of the position) with
the momentum operator.
Notice that the Hamiltonian for a set of harmonic oscillators, each having unit mass, is
1
Hh.o. =
X
(p2 2
2 n + ωnq2
n)
n
with q d
n, pn = qn
the position and momentum of each oscillator.
dt
10.2 Quantization of the e.m. field
Given the Hamiltonian we found above, we can associate the energy 1
(p2
n + ω2
q2
n) to each mode. We thus make the
2
identification of fn with an equivalent position:
fn
Qn =
2ωn
√
π
and then proceed to quantize this effective position, associating an operator to the position Qn:
Q̂n =
r
~
(a†
)
2 n + an
ωn
We can also associate an operator to the normal mode coefficients fn:
ˆ
fn = 2πω ~
n (a†
n + an)
33
The factor 4π is present because I am using cgs unit
p
s, in SI units the energy density is ǫ0
(E2
+ c2
B2
).
2
95

4. Notice that fn(t) is a function of time, so also the operators an(t) are (Heisenberg picture). The electric field is the
sum over this normal modes (notice that now the position is just a parameter, no longer an operator):
E(x, t) =
X p
2~πωn[a†
n(t) + an(t)]un(x)
n
Of course now the electric field is an operator field, that is, it is a QM operator that is defined at each space-time
point (x,t).
Notice that an equivalent formulation of the electric field in a finite volume V is given by defining in a slightly
different way the un(x) normal modes and writing:
E(x, t) =
X
r
2~πωn
[a†
n(t) + an(t)]un(x).
V
n
We already have calculated the evolution of the operator a and a†
. Each of the operator an evolves in the same way:
an(t) = an(0)e−iωnt
. This derives from the Heisenberg equation of motion dan
= i
~ [ , an(t)] = iωnan(t).
dt
The magnetic field can also be expressed in terms of the operators a
H −
n:
2π~
B(x, t) =
X
icn
r
[a†
n − an]
n
∇ )
ωn
× un(x
The strategy has been to use the known forms of the operators for a harmonic oscillator to deduce appropriate
dh (t)
operators for the e.m. field. Notice that we could have used the equation n
= cfn(t) to eliminate fn and write
dt
everything in terms of hn. This would have corresponded to identifying h
−
n with position and fn with momentum.
Since the Hamiltonian is totally symmetric in terms of momentum and position, the results are unchanged and we
can choose either formulations. In the case we chose, comparing the way in which the raising and lowering operators
enter in the E and B expressions with the way they enter the expressions for position and momentum, we may
say that, roughly speaking, the electric field is analogous to the position and the magnetic field is analogous to the
momentum of an oscillator.
10.2.1 Zero-Point Energy and the Casimir Force
The Hamiltonian operator for the e.m. field has the form of a harmonic oscillator for each mode of the field34
. As
we saw in a previous lecture, the lowest energy of a h.o. is 1
~ω. Since there are infinitely many modes of arbitrarily
2
high frequency in any finite volume, it follows that there should be an infinite zero-point energy in any volume of
space. Needless to say, this conclusion is unsatisfactory. In order to gain some appreciation for the magnitude of
the zero-point energy, we can calculate the zero-point energy in a rectangular cavity due to those field modes whose
frequency is less than some cutoff ωc. The mode functions un(x), solutions of the mode equation for a cavity of
dimensions Lx × Ly × Lz, have the vector components
un,α = Aα cos(kn,xrx) sin(kn,yry) sin(kn,zrz)
for {α, β, γ} = { ~
x, y, z} and permutations thereof. The mode un,α(x) are labeled by the wave-vector kn with compo-
nents: nαπ
kn,α = , nα
Lα
∈ N
and the frequency of the mode is ωn = k2 2
n,x + k2
n,y + kn,z. At least two of the integers must be nonzero, otherwise
the mode function would vanish identic
q
ally.
The amplitudes of the three components Aα are related by the divergence condition ∇ · ~
un(x) = 0, which requires
~ ~
that A · k = 0, from which it is clear that there are two linearly independent polarizations (directions of A) for each
34
See: Leslie E. Ballentine, “Quantum Mechanics A Modern Development”, World Scientific Publishing (1998). We follow
his presentation in this section.
96

5. k, and hence there are two independent modes for each set of positive integers (nx, ny, nz). If one of the integers is
zero, two of the components of u(x) will vanish, so there is only one mode in this exceptional case.
In this case the electric field can be written as:
E(x, t) =
X
(Eα + Eα
†
) =
X
êα
X
r
~ωn ~ ~
[a†
ei(kn·~
r−ωt)
+ a ei(kn·~
r−ωt)
n ]
2ǫ0V n
α=1,2 α=1,2 n
Here V = LxLyLz is the volume of the cavity. Notice that the electric field associated with a single photon of
frequency ωn is
~ω
E
r
2π n
n =
V
This energy is a figure of merit for any phenomena relying on atomic interactions with a vacuum field, for example,
cavity quantum electrodynamics. In fact, En may be estimated by equating the quantum mechanical energy of a
photon ~ωn with its classical energy 1
dV (E2
+ B2
).
2
Going back to the calculation of the en
R
ergy density, if the dimensions of the cavity are large, the allowed values of k
approximate a continuum, and the density of modes in the positive octant of k space is ρ(k) = 2V/π3
(the factor 2
comes from the two possible polarizations). The zero-point energy density for all modes of frequency less that ωc is
then given by
k
2 c
1 2 1
E = ω ≈
Z
d3 1
~
0 k kρ(k) ~ω(k)
V 2 V 8 2
k=1
where we sum over all positive k (in the firs
X
t sum) and multiply by the number of possible polarizations (2). The
sum is then approximated by an integral over the positive octant (hence the 1/8 factor). Using ω(k) = kc (and
d3
k = 4πk2
dk), we obtain
2 2V 4π
E0 =
V π3 8
Z kc
1 kc
dk ~k3 c~ ~
c = dkk3 ck4
= c
2 2π2
k=0
Z
k=0 8π2
where we set the cutoff wave-vector kc = ωc/c. The factor k4
c indicates that this energy density is dominated by
the high-frequency, short-wavelength modes. Taking a minimum wavelength of λ 6
c = 2π/k = 0.4 10−
m, so as to
include the visible light spectrum, yields a zero-point energy density of 23 J/m3
. This may be com
×
pared with energy
density produced by a 100 W light bulb at a distance of 1 m, which is 2.7 10−8
J/m3
. Of course it is impossible
to extract any of the zero-point energy, since it is the minimum possible e
×
nergy of the field, and so our inability
to perceive that large energy density is not incompatible with its existence. Indeed, since most experiments detect
only energy differences, and not absolute energies, it is often suggested that the troublesome zero-point energy of
the field should simply be omitted. One might even think that this energy is only a constant background to every
experimental situation, and that, as such, it has no observable consequences. On the contrary, the vacuum energy
has direct measurable consequences, among which the Casimir effect is the most prominent one.
In 1948 H. B. G. Casimir showed that two electrically neutral, perfectly conducting plates, placed parallel in vacuum,
modify the vacuum energy density with respect to the unperturbed vacuum. The energy density varies with the
separation between the mirrors and thus constitutes a force between them, which scales with the inverse of the
forth power of the mirrors separation. The Casimir force is a small but well measurable quantity. It is a remarkable
macroscopic manifestation of a quantum effect and it gives the main contribution
bodies for distances beyond 100nm.
We consider a large cavity of dimensions V = L3
bounded by conducting walls
(see figure). A conducting plate is inserted at a distance R from one of the yz
faces (R ≪ L). The new boundary condition at x = R alters the energy (or
frequency) of each field mode. Following Casimir, we shall calculate the energy
shift as a function of R. Let WX denote the electromagnetic energy within a
cavity whose length in the x direction is X. The change in the energy due to the
insertion of the plate at x = R will be
∆W = (WR + WL R) W
− − L
Each of these three terms is infinite, but the difference will turn out to be finite.
Each mode has a zero-point energy of 1
~kc. But while we can take the continuous
2
97
to the forces between macroscopic
WL WR WL-R
L
R
Fig. 14: Geometry of Casimir Effect

6. approximation in calculating WL and WL R, for WR we have to keep the discrete
−
sum in the x direction (if R is small enough). With some calculations (see Ballentine) we find that the change in
energy is
π2
L2
∆W = −~c
720 R3
When varying the position R, an attractive force (minus sign) is created between the conducting plates, equal to
∂∆W π2
L2
F = − = −~c
∂R 240 R4
2
The force per unit area (pressure) is then P = π ~c
4 . This is the so-called Casimir force. This force is very difficult
240 R
to measure. The surfaces must be flat and cle
−
an, and free from any electrostatic charge. However, there has been
measurements of the Casimir effect, since the experiment by by Sparnaay (1958).
The availability of experimental set-ups that allow accurate measurements of surface forces between macroscopic
objects at submicron separations has recently stimulated a renewed interest in the Casimir effect and in its possible
applications to micro- and nanotechnology. The Casimir force is highly versatile and changing materials and shape
of the boundaries modifies its strength and even its sign. Modifying strength and even sign of the Casimir force has
great potential in providing a means for indirect force transmission in nanoscale machines, which is at present not
achievable without damaging the components. A contactless method would represent a breakthrough in the future
development of nanomachines. More generally, a deeper knowledge of the Casimir force and Casimir torque could
provide new insights and design alternatives in the fabrications of micro- and nanoelectromechanical-systems (MEMS
and NEMS). Another strong motivation comes from the need to make advantage of the unique properties of Carbon
Nanotubes in nanotechnology.
Measuring the Casimir force is also important from a fundamental standpoint as it probes the most fundamental
physical system, that is, the quantum vacuum. Furthermore, it is a powerful experimental method for providing
constraints on the parameters of a Yukawa-type modification to the gravitational interaction or on forces predicted
by supergravity and string theory.
98

7. 10.3 Quantization of the e.m. field in the Coulomb gauge
The quantization procedure and resulting interactions detailed above may appear quite general, but in fact we made
an important assumption at the very beginning which will limit their applicability: we considered only the situation
~
with no sources, so we implicitly treated only transverse fields where ∇·E = 0. Longitudinal fields result from charge
distributions ρ and they do not satisfy a wave equation. By considering only transverse fields, however, we have
further avoided the issue of gauge. Since a transverse electric field ET satisfies the wave equation, we were able to
directly quantize it without intermediate recourse to the vector potential A and thus we never encountered a choice
~
of gauge. In fact, the procedure can be viewed as corresponding to an implicit choice of gauge φ = 0, A = 0 that
corresponds to a Lorentz gauge.
∇ ·
A more general approach may use the canonical Hamiltonian for a particle of mass m and charge q in an electro-
magnetic field. In this approach, the particle momentum p is replaced by the canonical momentum p qA/c, so the
Hamiltonian contains terms like H ∼ (p qA/c)2
/2m. In this case, it is still possible to write a wave eq
−
uation for the
potentials. Then the potential are quant
−
ized and for an appropriate choice of gauge we find again the same results.
Specifically, for an appropriate choice of gauge, the p · A terms imply the dipole interaction E · d that we will use in
the following.
Within the Coulomb gauge, the vector potential obeys the wave equation
∂2
A
∂t2
− c2
∇2
A = 0
Taking furthermore periodic boundary conditions in a box of volume V = L3
the quantized electromagnetic field in
the Heisenberg picture is:
2π~
~ ~ ~
A(t, x) =
X X
r h
a i(ωkt k ~
x) i(ωkt k ~
x)
kαe− − ·
+ ak
†
e êα(
V ω α
− ·
k)
k
α=1,2 k
i
The field can then be written in terms of the potential as E = −∂A
and we find the similar result as before:
∂t
E(t, x) =
X X
r
2π~ωk
h
~ ~
a i(ωkt k ~
x) i(ωkt k ~
x)
kαe− − ·
a†
kαe − ·
êα(k)
V
α=1,2 k
−
i
and
X X
r
2π~ωk
B(t, x) =
V
h
~
akαe−i(ωkt−k·~
x)
− a i
†
k
−~
αe (ωkt k·~
x)
(k êα(k))
α=1,2 k
i
×
10.4 States of the e.m. field
Because of the analogies of the e.m. with a set of harmonic oscillators, we can apply the knowledge of the h.o. states
to describe the states of the e.m. field. Specifically, we will investigate number states and coherent states.
10.4.1 Photon number eigenstates
We can define number states for each mode of the e.m. field. The Hamiltonian for a single mode is given by m =
~ωm(a†
mam + 1
) with eigenvectors |nmi. The state representing many modes is then given by
H
2
|n1, n2, . . .i = |n1i ⊗ |n2i ⊗ · · · = |~
ni
Therefore the mth
mode of this state is described as containing nm photons. These elementary excitations of the e.m.
field behave in many respects like particles, carrying energy and momentum. However, the analogy is incomplete,
and it is not possible to replace the e.m. field by a gas of photons.
99

8. In a state with definite photon numbers, the electric and magnetic fields are indefinite and fluctuating. The probability
distributions for the electric and magnetic fields in such a state are analogous to the distributions for the position
and momentum of an oscillator in an energy eigenstate. Thus we have for the expectation value of the electric field
operator:
hE(x, t)i = h~
n| 2~πωm[a†
m(t) + am(t)]um(x)
m
|~
ni = 0
However, the second moment is non-zero:
X p
|E(x, t)|2
= 2π~
X √
ωpωm
[a†
p(t) + ap(t)][a†
m(t) + am(t)]
~
up(x) )
p
· ~
um(x
,m
= 2π~ ωm|um|2
[a†
m(t) + a ( 2
m t)]2
= 2π~ ωm um
m m
| (x)| (2nm + 1)
The sum over all modes is infin
X
ite. This div
ergence problem c
an often
X
be circumvented (but not solved) by recognizing
that a particular experiment will effectively couple to the EM field only over some finite bandwidth, thus we can set
cut-offs on the number of modes considered.
Notice that we can as well calculate ∆B for the magnetic field, to find the same expression.
10.4.2 Coherent states
A coherent state of the e.m. field is obtained by specifying a coherent state for each of the mode oscillators of the
field. Thus the coherent state vector will have the form
|α
~i = |α1α2 . . .i = |α1i ⊗ |α2i ⊗ . . .
It is parameterized by a denumberably infinite sequence of complex numbers. We now want to calculate the evolution
of the electric field for a coherent state. In the Heisenberg picture it is:
E(x, t) =
X p
2 t
~πω [a† m
meiω
m + a t
me−iωm
]um(x)
m
then, taking the expectation value we find:
hE(x, t)i =
X p
2~πωm[αm
∗
eiωmt
+ αme−iωmt
]um(x)
m
This is exactly the same form as a normal mode expansion of a classical solution of Maxwell’s equations, with the
parameter αm representing the amplitude of a classical field mode. In spite of this similarity, a coherent state of the
quantized EM field is not equivalent to a classical field, although it does give the closest possible quantum operator,
in terms of its expectation value. Even if the average field is equivalent to the classical field, there are still the
characteristic quantum fluctuations. A coherent state provides a good description of the e.m. field produced by a
laser. Most ordinary light sources emit states of the e.m. field that are very close to a coherent state (lasers), or to a
statistical mixture of coherent states (classical sources).
A. Fluctuations
We calculate the fluctuations for a single mode ∆Em. From
2
|Em
2
~
|2
= hαm| Em · Em |αmi we obtain ∆E2
m =
2π ωm|~
um(x)| . Indeed, from
(am + a†
m) we obtain:
hαm| (am
†
)2
+ a2
m + a†
mam + ama†
m |α
mi 2π~ω|um(x)|2
=
1 + ((α∗
m)2
+ α2
m + 2α∗
mαm)
2π~ω|um(x)|2
while we have
am + a†
m
= (αm + α∗
m)
√
2π~ωum(x), so that we obtain
(am + a†
m)2
−
am + a†
m
2
= 2π~ωm|~
um(x)|2
This is independent of αm, and is equal to the mean square fluctuation in the ground state. The Heisenberg inequality
is therefore saturated when the field is in a coherent state,
100

9. B. Photon statistics
The photon number distribution for each mode in a coherent state is obtained as for the h.o. The probability of
finding a total of n photons in the field mode is governed by the Poisson distribution.
The probability of finding n photons in the mode m is Pα(nm) = |hnm|αi|2
. Using the expansion of the coherent
α 2n 2
state in terms of the number states that we found for the h.o., we obtain Pα(nm) = | m|
e−|αm|
. This is a Poisson
n!
| | n n
distribution, with parameter α 2
m . Thus we have hnmi = |αm|2
, so that we can rewrite the pdf as P(n) = h i
e
n!
−hni
.
2
From the known properties of the Poisson distribution, we also find ∆n2
= n2
− hni = hni.
In particular we have the well-know shot-noise scaling ∆n
= 1/
u
p
hni (i.e. the fl ctuations go to zero when there are
hni
many photons.)
10.4.3 Measurement Statistics
We saw in the previous section the photon number distribution for a coherent state. This corresponds to the exper-
imental situation in which we want to measure the number of photons in a field (such as laser light) which is well
approximated by a classical field and thus can be represented by a coherent state.
This is not the only type of measurement of the e.m. field that we might want to do. Two other common measurement
modalities are homodyne and heterodyne detection35
.
Homodyne detection corresponds to the measurement of one quadrature amplitude. In practice, one mixes the e.m.
field with a local oscillator at with a fixed frequency ω (same as the field frequency) before collecting the signal with
Fig. 15: Homodyne detection scheme and measurement statistics of the first three photon number eigenstates.
a photon counting detector.
Thus the measurement corresponds to the observable Oho = |α1i hα1| (or Oho = |α2i hα2|, depending on the phase of
the local oscillator), where |α1,2i are the eigenstates of the quadrature operators a1 = 1
(a + a†
) and a2 = i
(a†
− a).
2 2
The measurement statistics for a number state |mi is thus:
(a†
)n
ω 1
Pm(α1) = |hα1|mi|2
= hα1| |ni =
r
H2 2
m(α1/2)e−α2
1/
, hOho
n! π
i =
~
hm| Ohe |mi = 0, h∆Ohoi =
2
2
where Hn is the nth
Hermite polynomial and h∆Oi =
q
hO2i − hOi . We note that these results correspond to what
we had found for the x operator in the case of the quantum harmonic oscillator.
Heterodyne detection corresponds to the simultaneous measurement of the two quadratures of a field. Operationally,
one mixes the e.m. field with a local oscillator of frequency ω, modulated at the Intermediate Frequency ωIF ; the
35
We follow here the presentation in Prof. Yamamoto’s Lectures
101
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10. Fig. 16: Heterodyne detection scheme and measurement statistics of the first three photon number eigenstates.
signal, after collection, is demodulated by mixing it with sin(ωIF ) and cos(ωIF t). Thus the measurement corresponds
to the observable Ohe = |αi hα|. The measurement statistics for a number state |mi is thus:
α|2
e−|
|α|2n
P (α) = |hα|mi|2
= , hO i = hm| O |mi = |α|2
, h∆O 2
m he he he =
n
i |α
!
|
(note that of course this is equivalent to the case were we measured a number state for a coherent state). The
measurement statistics for a coherent state |βi, would be
Pβ(α) = |h |
2
α βi|2
= e−|β−α|
Fig. 17: Photon counting detection scheme and measurement statistics of the first three photon number eigenstates.
For comparison, photon counting is of course the measurement of the observable On = |ni hn|, with statistics for a
number state |mi
Pm(n) = |hn|mi|2
= δm,n, hOni = hm| On |mi = δm,n, h∆Oni = 0
10.5 Atomic interactions with the quantized field
Let us consider the interaction of isolated neutral atoms with optical fields. Such atoms alone have no net charge and
~
no permanent electric dipole moment. In an electric field E associated, e.g., with an electromagnetic wave, the atoms
102
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11. ~
do develop an electric dipole moment d which can then interact with the electric field with an interaction energy V
given by
~
V = d · ~
E
We have already treated a similar case in a semiclassical way, although we were interested in the interaction with
a magnetic field. The semi-classical treatment of this interaction, is quite similar: we treat the atom quantum
~
mechanically and therefore consider d as an operator, but treat the electromagnetic field classically and so consider
E as a vector. We can write the dipole moment operator as
~
d = e~
r =
X
k,h
|ki hk| ~
d |hi hh|
where {|ki} forms a complete basis. Transitions are only possible between states with different h and k:
~
dh,k = h ~
h| d |ki = 0 iif k = h
and we will consider for simplicity a two-level atom:
~
d = | ~ ~
0ih1|d01 + |1ih0|d10
~ ~
Let’s first consider a single mode classical electromagnetic field, given by E = Ee−iωt
+E∗
eiωt
. The full semi-classical
(sc) Hamiltonian is then:
1
H ~ | ih | − | ih | − | ih |~ ~ ~ i
sc = ω0( 1 1 0 0 ) ( 0 1 d ωt ~ iωt
01 + |1ih0|d10) · (Ee−
+ E∗
e )
2
~
If we assume {d10, E} ∈ R, we can rewrite this as
1
H ~
sc = ~ ~
ω0σz
2
− 2σxd10 · E cos(ωt)
Notice the correspondence with the spin Hamiltonian Hspin = Ωσz +B cos(ωt)σx describing the interaction of a spin
with a time-varying, classical magnetic field.
We can now go into the interaction frame defined by the Hamiltonian H0 = 1
ω0σz. Then we have:
2
H̃ = −(|0ih1|~
d eiω0t ~
+ | · (E
~
1ih0|d e−iω0t
) e−iωt ~ ω
c 01 10 + E∗
ei t
s )
On resonance (ω0 = ω) we retain only time-independent contributions to the Hamiltonian (RWA), then
H̃sc ≈ −(|1ih0|d E + |0ih1|d∗
E∗
)
Assuming for example that d E is real, we obtain an Hamiltonian −d Eσx, in perfect analogy with the TLS already
studied. (A more general choice of d E just gives an Hamiltonian at some angle in the xy plane).
Now let us consider a full quantum-mechanical treatment of this problem. The interaction between an atom and a
quantized field appears much the same as the semiclassical interaction. Starting with the dipole Hamiltonian for a
two-level atom, we replace E by the corresponding operator, obtaining the interaction Hamiltonian
V = −~
d · ~
E = −
X
~ ~
(E ~
α + Eα
† ~
) · (d
α
|1ih0| + d∗
|0ih1|)
X X
r
2π~ω ~
= −
m
[a† i k
me− ~
km·~
r
+ a m
mei ·~
r
](dα
V
α m
|1ih0| + d∗
α|0ih1|)
(where α is the polarization and m the mode). As in the semiclassical analysis, the Hamiltonian contains four terms,
which now have a clearer physical picture:
6 6
103

12. a†
m|0ih1| Atom decays from |1i → |0i and emits a photon (in the mth
mode).
a |1ih0| Atom is excited from |0i → |1i and absorbs a photon (from the mth
m mode).
a†
m|1ih0| Atom is excited from |0i → |1i and emits a photon (in the mth
mode).
am|0ih1| Atom decays from |1i → |0i and absorbs a photon (from the mth
mode).
For photons near resonance with the atomic transition, the first two processes conserve energy; the second two
processes do not conserve energy, and intuition suggests that they may be neglected. In fact, there is a direct
correspondence between the RWA and energy conservation: the second two processes are precisely those fast-rotating
terms we disregarded previously.
Consider the total Hamiltonian :
~ 1
H = H0 + V = ω ~
0σz +
X
ωm
m
a†
mam +
2 2
+ V
If we go to the interaction frame defined by the Hamiltonian 0, each mode acquires a time dependence e±iωmt
while
the atom acquires a time dependence e±iω0t
:
H
X
E
~
( a†
e− ~
ikm·~
r
eiωmt
+ E∗
a eikm r
m m m
·~
m e−iωmt
) · (e+iω0t
dα|1ih0| + d∗
αe−iω0at
|0ih1|)
where E =
q
2π~ωm
m . Thus the time-dependent factors acquired are
V
a†
m|0ih1| → a†
m|0ih1|e+i(ω0−ωm)t
am|1ih0| → am|1ih0|e−i(ω0−ωm)t
a†
m|1ih0| → a†
m|1ih0|ei(ω0+ωm)t
am|0ih1| → am|0ih1|e−i(ω0+ωm)t
For frequencies ωm near resonance ωm ≈ ω0, we only retains the first two terms.
~
Then, defining the single-photon Rabi frequency, g = −d ~
α 2
m ~
q
π ωm r
α eikm
,
·~
, the Hamiltonian in the interaction
V
picture and in the RWA approximation is
H =
X
~ gm,αam|1ih0| + gm
∗
,αa†
m
m
|0ih1|
From now on we assume an e.m. with a single mode (or we assume that only one mode is on resonance). We can
write a general state as |ψi =
P
n αn(t) |1ni + βn(t) |0ni, where n = nm is a state of the given mode m we retain
and here I will call the Rabi frequency for the mode of interest g
|
.
i
The
|
evo
i
lution is given by:
i~
X
˙
α̇n |1ni + βn |0ni = ~
X
g
αnσ a†
|1ni + βnσ+a |0n
−
n n
i
= ~ g αn
√
n + 1 |0, n + 1
n
i + βn
√
n |1, n − 1i
We then project these equations on
X
h1n| and h0n
|:
i~α̇n = ~gβn+1(t)
√
n + 1
i~β̇n = ~gαn−1(t)
√
n
to obtain a set of equations:
α̇n = −ig
√
n + 1βn+1
β̇n+1 = −ig
√
n + 1αn
This is a closed system of differential equations and we can solve for αn, βn+1.
We consider a more general case, where the field-atom are not exactly on resonance. We define ∆ = 1
(ω0 ω), where
2
ω = ωm for the mode considered. Then the Hamiltonian is:
−
H = ~ g a|1ih0| + g∗
a†
|0ih1| + ~∆ (|1ih1| − |0ih0|)
104

13. We can assume that initially the atom is in the the excited state |1i (that is, βn(0) = 0, ∀n). Then we have:
i∆t/2
Ωnt
i∆
αn(t) = αn(0)e cos
2
− sin
Ωn
Ωnt
2
ig
√
n
β i∆t/2 2 + 1 Ωnt
n(t) = −αn(0)e−
sin
Ωn
2
with Ω2
n = ∆2
+ 4g2
(n + 1). If initially there is no field (i.e. the e.m. field is in the vacuum state) and the atom is in
the excited state, then α0(0) = 1, while αn(0) = 0 ∀n = 0. Then there are only two components that are different
than zero:
α i∆t/2 Ω0t i∆ Ω0t
0(t) = e
cos
− sin
2 ∆2 + 4g2
2
#
2ig Ω0t
β0(t) = −e−i∆t/2
p
sin
∆2 + 4g2
2
or on resonance (∆ = 0)
p
gt
h1, n = 0|ψ(t)i = α0(t) = cos
2
gt
h0, n = 0|ψ(t)i = β0(t) = −i sin
2
Thus, even in the absence of field, it is possible to make the transition from the ground to the excited state! In the
semiclassical case (where the field is treated as classical) we would have no transition at all. These are called Rabi
vacuum oscillations.
6
105

14. 106

15. MIT OpenCourseWare
http://ocw.mit.edu
22.51 Quantum Theory of Radiation Interactions
Fall 2012
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